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Which is more likely to dissolve in an acidic solution, silver sulfide or silver chloride? Why?

Short Answer

Expert verified
Silver sulfide (Ag2S) is more likely to dissolve in an acidic solution because, in acidic environments, sulfide ions combine with H⁺ ions to form H2S gas, decreasing the concentration of S²⁻ ions and allowing more Ag2S to dissolve, while silver chloride (AgCl) does not react with H⁺ ions and its dissolution remains unaffected.

Step by step solution

01

Write down the solubility equilibria for silver sulfide and silver chloride

First, let's write down the solubility equilibrium reactions for each compound. For silver sulfide (Ag2S): \[Ag2S(s) \rightleftharpoons 2Ag^+(aq) + S^{2-}(aq)\] For silver chloride (AgCl): \[AgCl(s) \rightleftharpoons Ag^{+}(aq) + Cl^{-}(aq)\]
02

Study the behavior of each compound in acidic environments

Now, let's study how each compound behaves in acidic environments by writing the reaction equation of the respective anions with H+ ions: For sulfide ion (S²⁻): \[S^{2-}(aq) + H^+(aq) \rightleftharpoons HS^-(aq)\] \[HS^-(aq) + H^+(aq) \rightleftharpoons H2S(g)\] For chloride ion (Cl⁻): \[Cl^-(aq) + H^+(aq) \rightarrow \text{No reaction}\] In acidic solutions, sulfide ions combine with H⁺ ions to form H2S gas which escapes from the solution, effectively removing sulfide ions from the solution. This reduces the concentration of S²⁻ ions, allowing more Ag2S to dissolve and establish equilibrium. On the other hand, chloride ions do not react with H⁺ ions in acidic environments, and thus their concentration is not affected.
03

Compare the solubility of silver sulfide and silver chloride in acidic environments

Based on the behavior of these compounds in acidic solutions, we can conclude the following: Silver sulfide (Ag2S) is more likely to dissolve in acidic solutions when compared to silver chloride (AgCl). This is because, in acidic environments, sulfide ions combine with H⁺ ions to form H2S gas, which escapes the solution, decreasing the concentration of S²⁻ ions and allowing more Ag2S to dissolve to re-establish equilibrium. In contrast, silver chloride (AgCl) does not react with H⁺ ions, and its dissolution is not influenced by the acidity of the solution. So, the answer is: Silver sulfide (Ag2S) is more likely to dissolve in an acidic solution. The reason is that in acidic environments, sulfide ions combine with H⁺ ions to form H2S gas and decrease the concentration of S²⁻ ions, allowing more Ag2S to dissolve.

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Most popular questions from this chapter

Nitrate salts are generally considered to be soluble salts. One of the least soluble nitrate salts is barium nitrate. Approximately 15 g of \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}\) will dissolve per liter of solution. Calculate the \(K_{\mathrm{sp}}\) value for barium nitrate.

Solutions of sodium thiosulfate are used to dissolve unexposed \(\operatorname{AgBr}\left(K_{\mathrm{sp}}=5.0 \times 10^{-13}\right)\) in the developing process for black-and-white film. What mass of \(\mathrm{AgBr}\) can dissolve in \(1.00 \mathrm{L}\) of \(0.500 M\) \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3} ? \mathrm{Ag}^{+}\) reacts with \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) to form a complex ion: $$\begin{aligned} \mathrm{Ag}^{+}(a q)+2 \mathrm{S}_{2} \mathrm{O}_{3}^{2-}(a q) & \rightleftharpoons \mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}^{3-}(a q) \\ K &=2.9 \times 10^{13} \end{aligned}$$

Calcium oxalate \(\left(\mathrm{CaC}_{2} \mathrm{O}_{4}\right)\) is relatively insoluble in water \(\left(K_{\mathrm{sp}}=2 \times 10^{-9}\right) .\) However, calcium oxalate is more soluble in acidic solution. How much more soluble is calcium oxalate in 0.10\(M \mathrm{H}^{+}\) than in pure water? In pure water, ignore the basic properties of \(\mathrm{C}_{2} \mathrm{O}_{4}^{2-}.\)

Solubility is an equilibrium position, whereas \(K_{\mathrm{sp}}\) is an equilibrium constant. Explain the difference.

The salt MX has a solubility of \(3.17 \times 10^{-8} \mathrm{mol} / \mathrm{L}\) in a solution with \(\mathrm{pH}=0.000 .\) If \(K_{\mathrm{a}}\) for \(\mathrm{HX}\) is \(1.00 \times 10^{-15}\) , calculate the \(K_{\mathrm{sp}}\) value for \(\mathrm{MX}\) .

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